Governing equations on integral form: Difference between revisions

[[Category:Compressible flow]]

[[Category:Governing equations]]

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\noindent The governing equations stems from mass conservation, conservation of momentum and conservation of energy

\subsection{The Continuity Equation}

 

\noindent The net massflow into the control volume $\Omega$ in Fig. \ref{fig:generic:cv} is obtained by integrating mass flux over the control volume surface $\partial \Omega$\\

\[-\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS\]\\

\noindent Now, let's consider a small infinitesimal volume $d\mathscr{V}$ inside $\Omega$. The mass of $d\mathscr{V}$ is $\rho d\mathscr{V}$. Thus, the mass enclosed within $\Omega$ can be calculated as\\

\[\iiint_{\Omega} \rho d\mathscr{V}\]\\

\noindent The rate of change of mass within $\Omega$ is obtained as\\

\[\frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V}\]\\

\noindent Mass is conserved, which means that the rate of change of mass within $\Omega$ must equal the net flux over the control volume surface.\\

\[\frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V}=-\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS\]\\

or\\

\noindent which is the integral form of the continuity equation.\\

\subsection{The Momentum Equation}

\vspace*{1cm}

{\emph{''The time rate of change of momentum of a body equals the net force exerted on it''}}

\end{center}

\vspace*{1cm}

\[\frac{d}{dt}(m\mathbf{v})=\mathbf{F}\]\\

\noindent What type of forces do we have?\\

\noindent Body forces inside $\Omega$:\\

\[\iiint_{\Omega}\rho \mathbf{f}d\mathscr{V}\]\\

\noindent Surface force on $\partial \Omega$:\\

\[-\oiint_{\partial \Omega} p\mathbf{n}dS\]\\

\noindent Since we are considering inviscid flow, there are no shear forces and thus we have the net force as\\

\noindent The fluid flowing through $\Omega$ will carry momentum and the net flow of momentum out from $\Omega$ is calculated as\\

\[\oiint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n}dS)\mathbf{v}=\oiint_{\partial \Omega} (\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}dS\]\\

\noindent Integrated momentum inside $\Omega$\\

\[\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V}\]\\

\noindent Rate of change of momentum due to unsteady effects inside $\Omega$\\

\[\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V}\]\\

\noindent Combining the rate of change of momentum, the net momentum flux and the net forces we get\\

\noindent combining the surface integrals, we get\\

\noindent which is the momentum equation on integral form.\\

 

\newpage

 

\subsection{The Energy Equation}

 

\vspace*{1cm}

\begin{center}

{\emph{''Energy can be neither created nor destroyed; it can only change in form''}}

\end{center}

\vspace*{1cm}

\[E_1=\iiint_{\Omega} \dot{q}\rho d\mathscr{V}\]\\

\noindent The rate of work done on the fluid in $\Omega$ due to pressure forces is obtained from the pressure force term in the momentum equation.\\

\[E_{2_{pressure}}=-\oiint_{\partial \Omega}(p\mathbf{n}dS)\cdot\mathbf{v}=-\oiint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS\]\\

\noindent The rate of work done on the fluid in $\Omega$ due to body forces is\\

\[E_{2_{body\ forces}}=\iiint_{\Omega}(\rho\mathbf{f}d\mathscr{V})\cdot\mathbf{v}=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}\]\\

\[E_2=E_{2_{pressure}}+E_{2_{body\ forces}}=-\oiint_{\partial \Omega} p\mathbf{v}\cdot\mathbf{n}dS+\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}\]\\

\noindent The energy of the fluid per unit mass is the sum of internal energy $e$ (molecular energy) and the kinetic energy $V^2/2$ and the net energy flux over the control volume surface is calculated by the following integral\\

\[\oiint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right)\]\\

\noindent Analogous to mass and momentum, the total amount of energy of the fluid in $\Omega$ is calculated as\\

\[\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}\]\\

\noindent The time rate of change of the energy of the fluid in $\Omega$ is obtained as\\

\[\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}\]\\

\noindent Now, $E_3$ is obtained as the sum of the time rate of change of energy of the fluid in $\Omega$ and the net flux of energy carried by fluid passing the control volume surface.\\

\[E_3=\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}+\oiint_{\partial \Omega}(\rho \mathbf{v}\cdot\mathbf{n}dS)\left(e+\frac{V^2}{2}\right)\]\\

\noindent With all elements of the energy equation defined, we are now ready to finally compile the full equation\\

\frac{d}{dt}\iiint_{\Omega}\rho\left(e+\frac{V^2}{2}\right)d\mathscr{V}+\oiint_{\partial \Omega}\left[\rho\left(e+\frac{V^2}{2}\right)(\mathbf{v}\cdot\mathbf{n}) + p\mathbf{v}\cdot\mathbf{n}\right]dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}+\iiint_{\Omega} \dot{q}\rho d\mathscr{V}

\label{eq:governing:integral:energy}

\noindent The surface integral in the energy equation may be rewritten as\\

\noindent and with the definition of enthalpy $h=e+p/\rho$, we get\\

\noindent Furthermore, introducing total internal energy $e_o$ and total enthalpy $h_o$ defined as\\

\[e_o=e+\frac{1}{2}V^2\]\\

and\\

\[h_o=h+\frac{1}{2}V^2\]\\

\noindent the energy equation is written as\\

\subsection{Summary}

\noindent The integral form of the governing equations for inviscid compressible flow has been derived\\

\noindent Continuity:\\

\[\frac{d}{dt}\iiint_{\Omega} \rho d\mathscr{V}+\oiint_{\partial \Omega}\rho \mathbf{v}\cdot \mathbf{n} dS=0\]\\

\[\frac{d}{dt}\iiint_{\Omega} \rho \mathbf{v} d\mathscr{V}+\oiint_{\partial \Omega} \left[(\rho\mathbf{v}\cdot\mathbf{n})\mathbf{v}+p\mathbf{n}\right]dS=\iiint_{\Omega}\rho \mathbf{f}d\mathscr{V}\]\\

\noindent Energy:\\

\[\frac{d}{dt}\iiint_{\Omega}\rho e_o d\mathscr{V}+\oiint_{\partial \Omega}\rho h_o(\mathbf{v}\cdot\mathbf{n})dS=\iiint_{\Omega}\rho\mathbf{f}\cdot\mathbf{v}d\mathscr{V}+\iiint_{\Omega} \dot{q}\rho d\mathscr{V}\]